The number of solutions of the equation `x^(3)+x^(2)+4x+2sinx=0` in `0 le x le 2pi` is

Options

A. Zero
B. One
C. Two
D. Four

Answer

To find the number of solutions of the equation \( x^3 + x^2 + 4x + 2\sin x = 0 \) in the interval \( 0 \leq x \leq 2\pi \), we can follow these steps:

### Step 1: Define the function
Let
\[ f(x) = x^3 + x^2 + 4x + 2\sin x \]
We need to find the number of solutions to the equation \( f(x) = 0 \) in the interval \( [0, 2\pi] \).

### Step 2: Differentiate the function
Next, we differentiate \( f(x) \):
\[ f'(x) = 3x^2 + 2x + 4 + 2\cos x \]
Here, \( 3x^2 + 2x + 4 \) is a polynomial which is always positive for all \( x \) (since the discriminant \( b^2 - 4ac = 2^2 - 4 \cdot 3 \cdot 4 < 0 \)), and \( 2\cos x \) oscillates between -2 and 2.

### Step 3: Analyze the derivative
Since \( 3x^2 + 2x + 4 \) is always positive and \( 2\cos x \) can only decrease the value of \( f'(x) \) by at most 2, we can conclude that:
\[ f'(x) > 0 \quad \text{for all } x \in [0, 2\pi] \]
This means that \( f(x) \) is a strictly increasing function on the interval \( [0, 2\pi] \).

### Step 4: Evaluate the function at the endpoints
Now we evaluate \( f(x) \) at the endpoints of the interval:
- At \( x = 0 \):
\[ f(0) = 0^3 + 0^2 + 4 \cdot 0 + 2\sin(0) = 0 \]
- At \( x = 2\pi \):
\[ f(2\pi) = (2\pi)^3 + (2\pi)^2 + 4(2\pi) + 2\sin(2\pi) = 8\pi^3 + 4\pi^2 + 8\pi \]
Since \( 8\pi^3 + 4\pi^2 + 8\pi > 0 \), we have:
\[ f(2\pi) > 0 \]

### Step 5: Conclusion
Since \( f(x) \) is strictly increasing and \( f(0) = 0 \) while \( f(2\pi) > 0 \), there is exactly one point in the interval \( [0, 2\pi] \) where \( f(x) = 0 \).

Thus, the number of solutions of the equation \( x^3 + x^2 + 4x + 2\sin x = 0 \) in the interval \( 0 \leq x \leq 2\pi \) is:
\[ \boxed{1} \]